Question

A 1.989g sample of KClO3 is thermally decomposed losing 0.776g of oxygen. What is the experimental percent oxygen in KClO3?

Answer 1

You need to use the gravimetric factor to determine how much of the 1.989g is oxygen, simply use a ratio of the

[ (molecular mass of oxygen*3)/(Molecular mass of KClO3) ]*1.989g

Answer 2

you are simply figuring out the ratio of Oxygen present in the KClO3 molecule (just a percent calculation) using molecular mass then multiplying it by the original amount of the compound to figure out how much of the original mass is composed of oxygen

Answer 3

you need to pay attention because oxygen is a diatomic gas, so putting together a balanced reaction first is also important

Answer 4

You are probably right JFraser, but if you know how much oxygen is present in the original sample and how much oxygen is lost from that isn't a balanced reaction even necessary?

Answer 5

blah that should be is not isn't

Answer 6

I got .78 for what @Australopithecus said on his first post.

Answer 7

I know the true value of 03 is 39.1677g and the average for this ? is 39.0146?

Answer 8

Just to check my own work,

O 16*3 = 47

47/122 = 0.39

K 39

39/122 = 0.32

Cl 35

35/122 = 0.28

Summed up 0.99 or 99% (due to sketchy rounding)


But yeah to finish the problem you just have to take the percentage of

(Lost oxygen/Original oxygen)*100

Answer 9

and when you do gravimetric factor you always end up with what ever unit you are multiplying the ratio by, in this case you will get grams, it also works for moles

Answer 10

Lost oxygen = 0.776g / original oxygen=1.989?

Answer 11

KClO3 is 39% oxygen, 32% K, and 28% Cl

1.989 is the original mass of your KClO3 before it was decomposed.

Just take time and read what I wrote hopefully you will get it eventually

Answer 12

Ok, I did 0.776g-39.1677 all over 39.1677 and got 98%

Answer 13

That cant be right. I just dont get it

Answer 14

im so sorry i never learned this yet :(

Answer 15

it looks like you have been trying hard though :D

Answer 16

What do you need help with?

Answer 17

The ? posted at the top/

Answer 18

Does anyone know?

Answer 19

Of course we know.

Answer 20

You see, what you need to do is find the experimental percent of oxygen.

Answer 21

why were you subtracting 0.776g-39.1677?

If you have 1.989g of PURE KClO3, you simply need to figure out how much of that 1.989g is oxygen.

The experimental percentage oxygen, is the amount of oxygen that was generated from the original 1.989g of KClO3

So to figure this out you need to know how many grams of the 1.989g of your pure sample is made up of oxygen atoms. Which I showed you how to do.

Then simply take the percent oxygen obtained through decomposition divided by the original amount of oxygen in the 1.989g and then multiply it by 100

Answer 22

Do you follow?

Answer 23

sorry, "The experimental percentage oxygen, is the amount of oxygen that was generated from the original 1.989g of KClO3 through decomposition, relative to the amount of oxygen present in the original sample"

Answer 24

Yes, I think I get it now

Answer 25

Can you actually tell me what you got?

Answer 26

Site was down.

0.776 g/1.989 x100= 39.01 % oxygen


You're welcome.

Answer 27

Thank you.

Answer 28

I think my method is still valid, as mass and atomic ratio is conserved in reactions but you can do it another way that might seem more logical.

NOTE: this method wouldn't work where there was more then one product that contained oxygen.

The other method that is universal and will work everytime:

That is where you write out the decomposition of KClO3

Then figure out the total mass of O2 produced from the original reactants

then divide the mass of O2 obtained by the decomposition reaction and divide it by the theoretical amount of O2 obtained (which you figured out earlier using the decomposition formula and the original amount of reactant utilized). Then convert it to a percent by multiplying it by 100

This video might be helpful, https://www.youtube.com/watch?v=2TB8HWTs9D4

I don't have time to do the problem for you, if you want me to check your work I'm more than happy to do that for you. Just watch that video it has a problem identical to yours in it!